Answer
$(x,y)=(-\sqrt{5},-\sqrt{3})$, $(-\sqrt{5},\sqrt{3})$, $(\sqrt{5},-\sqrt{3})$, $(\sqrt{5},\sqrt{3})$
Work Step by Step
Adding the equations in the given system, \begin{align*}\require{cancel} x^2+y^2&=8 \\ x^2-y^2&=2 ,\end{align*} results to \begin{align*} 2x^2&=10 .\end{align*} Using the properties of equality, the equation $2x^2=10$ is equivalent to \begin{align*} \dfrac{\cancel2x^2}{\cancel2}&=\dfrac{10}{2} \\\\ x^2&=5 \\ x&=\pm\sqrt{5} &(\text{take the square root of both sides}) .\end{align*} Substituting the values of $x$ in the first equation results to \begin{align*} \begin{array}{l|r} \text{If }x=-\sqrt{5}: & \text{If }x=\sqrt{5}: \\\\ (-\sqrt{5})^2+y^2=8 & (-\sqrt{5})^2+y^2=8 \\ 5+y^2=8 & 5+y^2=8 \\ 5-5+y^2=8-5 & 5-5+y^2=8-5 \\ y^2=3 & y^2=3 \end{array} \end{align*} Taking the square root of both sides (Square Root Principle), then $y=\pm\sqrt{3}$.
Hence, the solutions of the given system are $(x,y)=(-\sqrt{5},-\sqrt{3})$, $(-\sqrt{5},\sqrt{3})$, $(\sqrt{5},-\sqrt{3})$, $(\sqrt{5},\sqrt{3})$.
